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I would like to mark a few blocks in a matrix and I followed directions from other similar questions but I get a weird TikZ behavior .. seems that I can not have more than one marked block in a matrix? What I would like to show is the stencil of a blocked Givens rotations to annihilate non-zero elements under the diagonal starting from some column, where I update a block (red color) over the diagonal and then can work on the trailing columns or row block on the right (green color).

UPDATE: Using the answer as input I implemented my final use case.

\usepackage{pgf}
\usepackage{tikz}
\usetikzlibrary{calc,fit,matrix,arrows,automata,positioning}

\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};}
\newcommand{\tikzdrawbox}[3][]{%
    \tikz[overlay,remember picture]{
    \draw[#3,#1]
      ($(left#2)+(-0.2em,0.9em)$) rectangle
      ($(right#2)+(0.2em,-0.3em)$);}
}

\newcommand\x{\times}

\begin{equation}\label{eq:blockedgivens}
\newcommand\y{\colorbox{myred}{$\times$}}
\newcommand\z{\colorbox{mygray}{$\times$}}
  \left(\begin{array}{ccccccccccccccc}
    \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 & \tikzmark{left1}\x & \x & \x & \x & \tikzmark{left3}\x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 & \z & \tikzmark{left2} \x \tikzmark{right1} & \x & \x & \x & \tikzmark{left4} \x & \x & \x & \x \tikzmark{right3}\ \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x \tikzmark{right2} & \x & \x & \x \tikzmark{right4} \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x \\
  \end{array}\right)
  \tikzdrawbox[thick]{1}{red}
  \tikzdrawbox[thick]{2}{red}
  \tikzdrawbox[thick]{3}{green}
  \tikzdrawbox[thick]{4}{green}
\end{equation}\

enter image description here

share|improve this question
    
Please add a few more lines and complete it to fully copy/pastable and hence compilable code. That would make the matter much easier to solve. Also don't forget to compile it twice. –  percusse Sep 1 '12 at 21:21
    
I know and I am sorry but it is not laziness from my side, it is a multi-page project adapted from a template and it is not really easy to turn into self contained snippets :( –  Giovanni Azua Sep 1 '12 at 21:25
2  
Related: How to highlight a single element in a matrix? –  Werner Sep 1 '12 at 22:17
1  
Also related: Highlight elements in the matrix –  Claudio Fiandrino Sep 2 '12 at 6:51
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2 Answers 2

up vote 7 down vote accepted

Each block need its own pair of marks; I changed the definition of \tikzdrawbox; the first mandatory argument is a number used for the marks; the second optional argument allows you to pass options to \draw and the optional argument allows you to specify some "correction" to the block dimensions to prevent some borders to overlap:

\documentclass{article}
\usepackage{pgf}
\usepackage{tikz}
\usetikzlibrary{calc,fit,matrix,arrows,automata,positioning}

\newcommand\x{\times}
\newcommand\y{\colorbox{red}{$\times$}}
\newcommand\z{\colorbox{gray}{$\times$}}


\newcommand{\tikzmark}[1]{\tikz[overlay,remember picture] \node (#1) {};}
\newcommand{\tikzdrawbox}[3][(0pt,0pt)]{%
    \tikz[overlay,remember picture]{
    \draw[#3]
      ($(left#2)+(-0.3em,0.9em) + #1$) rectangle
      ($(right#2)+(0.2em,-0.4em) - #1$);}
}

\begin{document}

\begin{equation}\label{eq:blockedgivens}
  \left(\begin{array}{ccccccccccccccc}
    \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 & \tikzmark{left1}\x & \x & \x & \x & \tikzmark{left3}\x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 & \z & \tikzmark{left2} \x \tikzmark{right1} & \x & \x & \x & \tikzmark{left4} \x & \x & \x & \x \tikzmark{right3} \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x \tikzmark{right2} & \x & \x & \x \tikzmark{right4} \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x & \x \\
     0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \z & \x \\
  \end{array}\right)
  \tikzdrawbox{1}{thick,red}
  \tikzdrawbox{2}{thick,red}
  \tikzdrawbox[(-1pt,2pt)]{3}{thick,green}
  \tikzdrawbox[(-4pt,2pt)]{4}{thick,green}
\end{equation}

\end{document}

enter image description here

share|improve this answer
    
you did it again :) –  Giovanni Azua Sep 1 '12 at 21:26
    
if I change the style of some elements the marked blocks shift right ... any ideas why? e.g. change the zero below the top left x to \z and you will see it shifts right. –  Giovanni Azua Sep 1 '12 at 21:49
    
I am sorry :( I have to unanswer it because doesn't work to implement my final use-case above. I get all block boundaries broken when I add all the elements I need, otherwise I have to create a new Question for the same thing and it is embarrassing :D –  Giovanni Azua Sep 1 '12 at 22:14
    
@GiovanniAzua What is the desired output you expect to achieve? I don't understand what do you mean with "broken boundaries". –  Gonzalo Medina Sep 1 '12 at 22:20
1  
@GiovanniAzua because you placed the marks to span five columns; take block 1, for example: initial mark is on column 4 and final mark is on column 8, so the block will span columns 4,5,6,7 and 8. –  Gonzalo Medina Sep 1 '12 at 22:40
show 5 more comments

If you don't have too many special entries(for which you can include more if cases), you can shorten the code a bit.

\documentclass{article}
\usepackage{tikz}

\begin{document}
\begin{equation}\label{eq:blockedgivens}
\left( 
\begin{tikzpicture}[baseline=(current bounding box.center)] 
\begin{scope}[xscale=0.6,yscale=0.4]
\foreach \x in {1,...,14}
    \foreach \y in {1,...,14}
    {
    \ifnum\x<\y
    \node (my-\y-\x) at (\x,-\y) {$0$};
    \else
    \node (my-\y-\x) at (\x,-\y) {$\times$};
    \fi
    }
\end{scope}
\draw[red] (my-4-4.north west) rectangle (my-8-8.south east);
\draw[red] (my-8-8.north west) rectangle (my-12-12.south east);
\end{tikzpicture}
\right)
\end{equation}
\end{document}

enter image description here

share|improve this answer
    
the thing is that the matrix array I understand well, but this for scripting of TikZ is still a big mystery for me. I have tried implementing several use-cases with it and always end up bumping to strange errors or dead ended by its incomprehensible markup. –  Giovanni Azua Sep 1 '12 at 21:58
1  
@GiovanniAzua Maybe if you can provide some small examples that would illustrate those cases we can have a look together. I didn't do much scripting I just put a node at each integer grid point and if the index is bigger than the other, I change its text (which is the standard if then else clause). And to reduce the size I've used a local scaling only valid inside the scope. –  percusse Sep 1 '12 at 22:01
    
Thank you percusse, where can I show you? hmm I will probably prepare a separate question otherwise this will become too OT w.r.t the OP –  Giovanni Azua Sep 1 '12 at 22:03
    
@GiovanniAzua Indeed, a new question (or more depending on the context) would also attract more wizards :) –  percusse Sep 1 '12 at 22:04
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